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-BARSIC- [3]
3 years ago
7

A wildlife researcher is tracking a flock of geese. the geese fly 3.5 km due west, then turn toward the north by 40 ∘ and fly an

other 4.5 km, what is the magnitude of their displacement

Physics
2 answers:
kaheart [24]3 years ago
4 0

To solve this problem, we can use the cosine formula for calculating the length of the displacement:

c^2 = a^2 + b^2 – 2 a b cos θ

 

where c is the displacement, a = 3.5 km, b = 4.5 km, and θ is the angle inside the triangle

 

Since the geeze turned 40° from west to north, so the angle inside the triangle must be:

θ = 180 – 40 = 140°

 

c^2 = 3.5^2 + 4.5^2 – 2 (3.5) (4.5) cos 140

c^2 = 56.63

c = 7.53 km

 

<span>So the magnitude of the displacement is 7.53 km</span>

emmasim [6.3K]3 years ago
4 0

The magnitude of their displacement is 7.5 km

<h3>Further explanation</h3>

Vector is a quantity that has a magnitude and direction.

A vector in a cartesian coordinate is represented by an arrow in which the slope of the arrow shows the direction of the vector and the length of the arrow shows the magnitude of the vector.

A position vector of a point is a vector drawn from the base point of the coordinates O (0,0) to that point.

The addition of two vectors can be done in the following ways:

\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}

A negative vector is a vector with the same magnitude but in opposite direction.

\overrightarrow {AB} = -\overrightarrow {BA}

Let's tackle the problem!

First, let's look at the picture in the attachment!

Let:

\overrightarrow{d_1} = -3.5 \widehat{i}

\overrightarrow{d_2} = -(4.5 \cos 40^o) \widehat{i} + (4.5 \sin 40^o) \widehat{j}

Then:

\overrightarrow{d} = \overrightarrow{d_1} + \overrightarrow{d_2}

\overrightarrow{d} = -3.5 \widehat{i} + -(4.5 \cos 40^o) \widehat{i} + (4.5 \sin 40^o) \widehat{j}

\overrightarrow{d} = -(3.5 + 4.5 \cos 40^o)\widehat{i} + (4.5 \sin 40^o) \widehat{j}

|\overrightarrow{d}| = \sqrt{ (3.5 + 4.5 \cos 40^o)^2 + (4.5 \sin 40^o)^2}

|\overrightarrow{d}| \approx 7.5 ~ km

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Magnitude of A Vector : brainly.com/question/2678571

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Vectors

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Vector , Scalar

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Given:

Speed of the speed 2.9 m/s

The direction in which the inertia change is moving = 43° north west.

Time taken to travel the distance = 30 minutes

According to the question that is moving in a Northwest direction with a velocity of 2.9 m per second.

If you want to find the distance it has moved in either North or west direction you will first have to find the total distance that kayak has moved in 30 minutes.

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The distance traveled in the west direction can be find using trigonometric ratio.

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The distance traveled in the west direction = 3.6 kilometers

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to
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Answer:

1) \Delta s=1000\ ft

2)  \Delta s'=998.11\ ft.s^{-1}

3) t\approx125\ s

t'\approx463.733\ s

Explanation:

Given:

width of river, w=500\ ft

speed of stream with respect to the ground, v_s=8\ ft.s^{-1}

speed of the swimmer with respect to water, v=4\ ft.s^{-1}

<u>Now the resultant of the two velocities perpendicular to each other:</u>

v_r=\sqrt{v^2+v_s^2}

v_r=\sqrt{4^2+8^2}

v_r=8.9442\ ft.s^{-1}

<u>Now the angle of the resultant velocity form the vertical:</u>

\tan\beta=\frac{v_s}{v}

\tan\beta=\frac{8}{4}

\beta=63.43^{\circ}

  • Now the distance swam by the swimmer in this direction be d.

so,

d.\cos\beta=w

d\times \cos\ 63.43=500

d=1118.034\ ft

Now the distance swept downward:

\Delta s=\sqrt{d^2-w^2}

\Delta s=\sqrt{1118.034^2-500^2}

\Delta s=1000\ ft

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

v'=v.\cos37

v'=4\times \cos37

v'=3.1945\ ft.s^{-1}

<u>Now the net effective speed of stream sweeping the swimmer:</u>

v_n=v_s-v'

v_n=8-3.1945

v_n=4.8055\ ft.s^{-1}

<u>The  component of swimmer's velocity heading directly towards the opposite bank:</u>

v'_r=v.\sin37

v'_r=4\sin37

v'_r=2.4073\ ft.s^{-1}

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

\tan\phi=\frac{v_n}{v'_r}

\tan\phi=\frac{4.8055}{2.4073}

\phi=63.39^{\circ}

  • Now let the distance swam in this direction be d'.

d'\times \cos\phi=w

d'=\frac{500}{\cos63.39}

d'=1116.344\ ft

<u>Now the distance swept downstream:</u>

\Delta s'=\sqrt{d'^2-w^2}

\Delta s'=\sqrt{1116.344^2-500^2}

\Delta s'=998.11\ ft.s^{-1}

3)

Time taken in crossing the rive in case 1:

t=\frac{d}{v_r}

t=\frac{1118.034}{8.9442}

t\approx125\ s

Time taken in crossing the rive in case 2:

t'=\frac{d'}{v'_r}

t'=\frac{1116.344}{2.4073}

t'\approx463.733\ s

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